Let $C$ be a locally finite $k$-linear abelian category. Here locally finite means that the home sets are finite dimensional vector space and every object of $C$ has finite length.
Suppose that each object of $C$ is projective. I want to prove that $C$ is semisimple.
Here us what I have thought:
Let $X$ be an object of $C$. Since it has finite length, there is a simple object $Y$ and monomorphism $i: Y \to X$.
Then we have an exact sequence $Y \to X \to X/Y \to 0$. Since $X/Y$ is projective, we have a unique morphism from $X/Y to X$ so that the sequence splits.
I don't know how to conclude that $X$ is a direct sum of $Y$ and $X/Y$ from here. I want to construct a morphism from $X$ to $Y$ as well.
Also, I have no idea how to prove that there are only finitely many simple objects. Could you give me a proof?
Suppose that $\mathbb{A}$ is an abelian category and that $0\to K \to A\to B\to 0$ is an exact sequence which splits. Let us write $\kappa: K\to A$, $\alpha:A\to B$ and $\beta : B\to A$ for the morphisms involved. By exactness we have that $\kappa$ is the kernel of $\alpha$, $\alpha$ is the cokernel of $\kappa$, and since the sequence splits $\alpha\beta =1_B$. Since $\kappa$ is the kernel of $\alpha$ and $\alpha(1_A - \beta\alpha) = \alpha - \alpha\beta\alpha = 0 $ it follows that there exists a unique morphism $\gamma : A\to K$ such that $\kappa\gamma = 1_A -\beta\alpha$. Note that since $\kappa$ is a monomorphism and $\kappa\gamma\kappa = (1_A-\beta\alpha)\kappa=\kappa=\kappa1_K$ it follows that $\gamma\kappa = 1_K$. Using that $\kappa$ is a monomorphism again and $\kappa \gamma \beta = (1_A -\beta\alpha) \beta= \beta - \beta\alpha\beta = \beta - \beta = 0 = \kappa 0$ we find that $\gamma \beta = 0$. We now have what Mac Lane calls a biproduct diagram (which is more often now called a direct sum diagram) i.e. we have objects $K$, $A$ and $B$, morphisms $\alpha:A\to B$, $\gamma:A\to K$, $\beta: B\to A$ and $\kappa: K\to A$ satisfying $\alpha \beta=1_B$, $\gamma\kappa=1_K$, $\alpha \kappa = 0$, $\gamma \beta =0$ and $\beta \alpha + \kappa\gamma = 1_A$. It is a relatively easy exercise to show that for such a diagram the diagram $X \leftarrow A \rightarrow B$ is a product and the diagram $K \rightarrow A \leftarrow B$ is a coproduct.